Respuesta :
Answer:
The bubble's volume = 2830 cm³
Explanation:
Step 1 : Data given
Depth of 20.0 m below the surface
Temperature is 5.00°C = 278 Kelvin
Volume = 0.90 cm³
Before it hits the ocean surface, its temperature is 20.0°C = 293 Kelvin
Density sea water = 1025 kg /m³
Step 2:
Let's assume that the pressure of the air in the bubble is the same as the pressurein the surrounding water.
Let's consider d as the deepth of the ocean and ρ is tge density of the water
p1 = p0 +pgd
⇒ p0 = atmospheric pressure
Since p1V1 = nRT1 we can calculate the numberof moles as:
n = p1V1/RT1 = (p0+pgf)*V1/RT1
⇒ V1 = the volume of the bubble at the bottom of the ocean
⇒ T1 = the temperature at the bottom of the ocean
At the surface of the ocean, the pressure = p0
The volume of the bubble is:
V2= nRT2/p0
V2=(T2/T1)*((p0+pgd)/p0) *V1
V2= (293/278) * ((101325 + 1025*9.81 *20)/101325)*0.9
V2 =1.054 * (101325+201.105)/101325)0.9
V2 = 1.054 * 1.002*0.9
V2 =2.83 L = 2830 cm³
The bubble's volume = 2830 cm³
Answer:
2.83 cm³
Explanation:
The ideal gas law states that the relationship between pressure (P), volume(V), temperature (T) and the moles (n) are constant:
PV/n*T = constant
So, assuming that the number of moles doesn't vary in this process:
P1*V1/T1 = P2*V2/T2
Where 1 is the state when the bubble under water, and 2 when it hits the surface. The pressure of the bubble is the same as its surroundings, so by the Stevin's theorem, inside a liquid:
P1 = Patm + ρ*g*h
Where Patm is the atmospheric pressure (101325 Pa), ρ is the density (1025 kg/m³), g is the gravity acceleration (9.8 m/s²), and h is the depth (20.0 m), so:
P1 = 101325 + 1025*9.8*20
P1 = 302225 Pa
P2 = Patm = 101325 Pa, T1 = 5.00°C = 278 K, T2 = 20°C = 293 K. So:
(302225*0.90)/278 = (101325*V2)/293
978.4263 = 345.8191V2
345.8191V2 = 978.4263
V2 = 2.83 cm³
